0
$\begingroup$

If I express the Dirac equation in the form of

$$i\hbar \frac{\partial}{\partial t} \psi_a(x) = \left(-i\hbar c(\alpha^j)_{ab}\partial _j + mc^2(\beta)_{ab}\right)\psi_b(x),$$

with the constraints

$$\{\alpha^j,\alpha^k\}_{ab} = 2\delta^{jk}\delta_{ab}, \qquad \{\alpha^j,\beta\}_{ab}=0, \qquad (\beta^2)_{ab} = \delta_{ab},$$

since the Gamma matrices are traceless and their eigenvalues are either +1 or -1, we can show that the Gamma matrices are of even dimension.

My question is: how can I show $2 \times 2$ matrices are not enough?

$\endgroup$
2
$\begingroup$

You need four linearly independent matrices. Write down the representation:

a + ib   c + id
e + if   g + ih

The traceless condition gives $a = -g$, $h=-b$. Now we have

a + ib   c + id
e + if  -a - ib

This has six unknowns (six possible independent matrices).

Then the eigenvalue conditions for the four matrices give four equations in six unknowns, which means that there is no linearly independent solution with $2\times2$ matrices, since that would require four degrees of freedom, but we only have two left.

Incidentally this question is often an exercise in books on quantum field theory or relativistic quantum mechanics.

$\endgroup$
  • 1
    $\begingroup$ Thank you! This simple reasoning is what I expected. Actually, I am reading a quantum field theory book. $\endgroup$ – Kevin Kwok Oct 20 '16 at 11:11
  • $\begingroup$ @KevinKwok - yes it is usually in those kinds of books. I had a look at the new edition of Mandl & Shaw (2010) and can't find it there so I seem to have been mistaken. $\endgroup$ – Suzu Hirose Oct 20 '16 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.